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baldymcgee6

  • 3 years ago

Partial Derivatives... posted below

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  1. baldymcgee6
    • 3 years ago
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    |dw:1351748975295:dw|

  2. hartnn
    • 3 years ago
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    partial derivative means u treat other variable as constant. u know this ?

  3. baldymcgee6
    • 3 years ago
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    Right, so we treat y as a constant in the first one..

  4. Dido525
    • 3 years ago
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    Which variable are you taking constant>?

  5. hartnn
    • 3 years ago
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    yes.

  6. baldymcgee6
    • 3 years ago
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    |dw:1351749134665:dw|

  7. hartnn
    • 3 years ago
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    absolutely! that was my next statement, that u need log diff. here.

  8. baldymcgee6
    • 3 years ago
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    In my experiences with log diff I always had a function in the form of y = f(x)^g(x).... So what do we do here?

  9. baldymcgee6
    • 3 years ago
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    do we write y = ......?

  10. hartnn
    • 3 years ago
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    taking log on both sides, u get log y = g(x) log f(x) then use product rule

  11. hartnn
    • 3 years ago
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    u'll also need a chain rule for this....

  12. baldymcgee6
    • 3 years ago
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    |dw:1351749340629:dw|

  13. hartnn
    • 3 years ago
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    oh dear! the y on left and on right are different, u should have used g or some other letter

  14. baldymcgee6
    • 3 years ago
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    oohhhhhh.. that's where I went wrong... this is off my last quiz :/

  15. baldymcgee6
    • 3 years ago
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    so it'll be g'/g on the left?

  16. hartnn
    • 3 years ago
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    yes, after doing product rule , u resubstitute for g

  17. baldymcgee6
    • 3 years ago
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    |dw:1351749800366:dw|

  18. hartnn
    • 3 years ago
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    thats right

  19. baldymcgee6
    • 3 years ago
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    |dw:1351749846652:dw|

  20. hartnn
    • 3 years ago
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    absolutely correct

  21. hartnn
    • 3 years ago
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    same for 2nd question, just replace x with y and y with x

  22. baldymcgee6
    • 3 years ago
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    right... I just messed up on calling the function y... messed with my head.

  23. baldymcgee6
    • 3 years ago
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    thanks again for the insight!

  24. baldymcgee6
    • 3 years ago
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    I'm not going to do the second derivative, I think I can get it... Thanks so much!

  25. baldymcgee6
    • 3 years ago
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    I mean, i'm not going to do it on here.. :)

  26. hartnn
    • 3 years ago
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    ok, welcome ^_^

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